Промежуточные таблицы истинности:¬X2:
X1∨(¬X2):
X1 | X2 | ¬X2 | X1∨(¬X2) |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 |
(X1∨(¬X2))∨X3:
X1 | X2 | X3 | ¬X2 | X1∨(¬X2) | (X1∨(¬X2))∨X3 |
0 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 1 |
X1∧(¬X2):
X1 | X2 | ¬X2 | X1∧(¬X2) |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 0 |
¬((X1∨(¬X2))∨X3):
X1 | X2 | X3 | ¬X2 | X1∨(¬X2) | (X1∨(¬X2))∨X3 | ¬((X1∨(¬X2))∨X3) |
0 | 0 | 0 | 1 | 1 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 0 |
(¬((X1∨(¬X2))∨X3))∨(X1∧(¬X2)):
X1 | X2 | X3 | ¬X2 | X1∨(¬X2) | (X1∨(¬X2))∨X3 | ¬((X1∨(¬X2))∨X3) | ¬X2 | X1∧(¬X2) | (¬((X1∨(¬X2))∨X3))∨(X1∧(¬X2)) |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
Y≡((¬((X1∨(¬X2))∨X3))∨(X1∧(¬X2))):
Y | X1 | X2 | X3 | ¬X2 | X1∨(¬X2) | (X1∨(¬X2))∨X3 | ¬((X1∨(¬X2))∨X3) | ¬X2 | X1∧(¬X2) | (¬((X1∨(¬X2))∨X3))∨(X1∧(¬X2)) | Y≡((¬((X1∨(¬X2))∨X3))∨(X1∧(¬X2))) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
Общая таблица истинности:
Y | X1 | X2 | X3 | ¬X2 | X1∨(¬X2) | (X1∨(¬X2))∨X3 | X1∧(¬X2) | ¬((X1∨(¬X2))∨X3) | (¬((X1∨(¬X2))∨X3))∨(X1∧(¬X2)) | Y≡¬(X1∨¬X2∨X3)∨(X1∧¬X2) |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
Y | X1 | X2 | X3 | F |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 |
F
сднф = ¬Y∧¬X1∧¬X2∧¬X3 ∨ ¬Y∧¬X1∧¬X2∧X3 ∨ ¬Y∧¬X1∧X2∧X3 ∨ ¬Y∧X1∧X2∧¬X3 ∨ ¬Y∧X1∧X2∧X3 ∨ Y∧¬X1∧X2∧¬X3 ∨ Y∧X1∧¬X2∧¬X3 ∨ Y∧X1∧¬X2∧X3
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
Y | X1 | X2 | X3 | F |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 |
F
скнф = (Y∨X1∨¬X2∨X3) ∧ (Y∨¬X1∨X2∨X3) ∧ (Y∨¬X1∨X2∨¬X3) ∧ (¬Y∨X1∨X2∨X3) ∧ (¬Y∨X1∨X2∨¬X3) ∧ (¬Y∨X1∨¬X2∨¬X3) ∧ (¬Y∨¬X1∨¬X2∨X3) ∧ (¬Y∨¬X1∨¬X2∨¬X3)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
Y | X1 | X2 | X3 | Fж |
0 | 0 | 0 | 0 | 1 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 1 | 0 |
Построим полином Жегалкина:
F
ж = C
0000 ⊕ C
1000∧Y ⊕ C
0100∧X1 ⊕ C
0010∧X2 ⊕ C
0001∧X3 ⊕ C
1100∧Y∧X1 ⊕ C
1010∧Y∧X2 ⊕ C
1001∧Y∧X3 ⊕ C
0110∧X1∧X2 ⊕ C
0101∧X1∧X3 ⊕ C
0011∧X2∧X3 ⊕ C
1110∧Y∧X1∧X2 ⊕ C
1101∧Y∧X1∧X3 ⊕ C
1011∧Y∧X2∧X3 ⊕ C
0111∧X1∧X2∧X3 ⊕ C
1111∧Y∧X1∧X2∧X3
Так как F
ж(0000) = 1, то С
0000 = 1.
Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F
ж(1000) = С
0000 ⊕ С
1000 = 0 => С
1000 = 1 ⊕ 0 = 1
F
ж(0100) = С
0000 ⊕ С
0100 = 0 => С
0100 = 1 ⊕ 0 = 1
F
ж(0010) = С
0000 ⊕ С
0010 = 0 => С
0010 = 1 ⊕ 0 = 1
F
ж(0001) = С
0000 ⊕ С
0001 = 1 => С
0001 = 1 ⊕ 1 = 0
F
ж(1100) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
1100 = 1 => С
1100 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 1 => С
1010 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
F
ж(1001) = С
0000 ⊕ С
1000 ⊕ С
0001 ⊕ С
1001 = 0 => С
1001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F
ж(0110) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0110 = 1 => С
0110 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
F
ж(0101) = С
0000 ⊕ С
0100 ⊕ С
0001 ⊕ С
0101 = 0 => С
0101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F
ж(0011) = С
0000 ⊕ С
0010 ⊕ С
0001 ⊕ С
0011 = 1 => С
0011 = 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F
ж(1110) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
1100 ⊕ С
1010 ⊕ С
0110 ⊕ С
1110 = 0 => С
1110 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1101) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0001 ⊕ С
1100 ⊕ С
1001 ⊕ С
0101 ⊕ С
1101 = 1 => С
1101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(1011) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
0001 ⊕ С
1010 ⊕ С
1001 ⊕ С
0011 ⊕ С
1011 = 0 => С
1011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(0111) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
0111 = 1 => С
0111 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
F
ж(1111) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
1100 ⊕ С
1010 ⊕ С
1001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
1110 ⊕ С
1101 ⊕ С
1011 ⊕ С
0111 ⊕ С
1111 = 0 => С
1111 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = 1 ⊕ Y ⊕ X1 ⊕ X2 ⊕ X2∧X3 ⊕ X1∧X2∧X3
Логическая схема, соответствующая полиному Жегалкина: